Rate Law

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Mary Becerra 2D
Posts: 53
Joined: Fri Sep 29, 2017 7:06 am

Rate Law

Postby Mary Becerra 2D » Thu Feb 22, 2018 2:14 pm

I understand that aA's rate would be (-1/a)*(dA/dt). Does this mean 1/a*A's rate is (-a)*(dA/dt)? Thank you!

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: Rate Law

Postby Danah Albaaj 1I » Thu Feb 22, 2018 2:34 pm

No, the answer would just be -1 because you still have to follow the formula and dividing the fractions will cancel out the value of a and leave you with -1.

Sandhya Rajkumar 1C
Posts: 50
Joined: Fri Jun 23, 2017 11:40 am

Re: Rate Law

Postby Sandhya Rajkumar 1C » Sun Feb 25, 2018 10:29 am

From my understanding of it, if your stoichiometric coefficient is now 1/a instead of a, you plug in 1/a to where you would plug in a. This leaves you with
(-1/(1/a))*(d[A]/dt) which simplifies to -a*(d[A]/dt)

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